Question

user_id  category     suburb      dated        walk_time
1        experience    US      2016-04-09        5
1        discovery     US      2016-04-09        5
1        experience    UK      2016-04-09        5
1        experience    AUS     2016-04-23        10
2        actions       IND     2016-04-15        2
2        actions       IND     2016-04-15        1
2        discovery     US      2016-04-21        2
3        discovery     FR      2016-04-12        3
3        Emotions      IND     2016-04-23        3
3        discovery     UK      2016-04-12        4
3        experience    IND     2016-04-12        3

我试图让每个用户最常用的类别，郊区，日期，walk_time

所以结果表将是

user_id  category     suburb      dated       walk_time
1        experience    US      2016-04-09        5
2        actions       IND     2016-04-15        2
3        discovery     IND     2016-04-12        3

我在这里尝试的查询是

select user_id,
       substring_index(group_concat(suburb order by cnt desc), ',', 1) as suburb_visited,
       substring_index(group_concat(category order by cct desc), ',', 1) as category_used,
       substring_index(group_concat(walk_time order by wct desc), ',', 1) as walked,
       substring_index(group_concat(dated order by nct desc), ',', 1) as dated_at
from (select user_id, suburb, count(*) as cnt,category, count(*) cct, walk_time, count(*) wct, dated,count(*) nct
      from temp_user_notes
      group by user_id, suburb,category,walk_time,dated
     ) upv
group by user_id;

Answer 1

SELECT user_id,
      (SELECT category FROM temp_user_notes t1
       WHERE t1.user_id = T.user_id
       GROUP BY category ORDER BY count(*) DESC LIMIT 1) as category,
      (SELECT suburb FROM temp_user_notes t2
       WHERE t2.user_id = T.user_id
       GROUP BY suburb ORDER BY count(*) DESC LIMIT 1) as suburb,
      (SELECT dated FROM temp_user_notes t3
       WHERE t3.user_id = T.user_id
       GROUP BY dated ORDER BY count(*) DESC LIMIT 1) as dated,
      (SELECT walk_time FROM temp_user_notes t4
       WHERE t4.user_id = T.user_id
       GROUP BY walk_time ORDER BY count(*) DESC LIMIT 1) as walk_time
FROM (SELECT DISTINCT user_id FROM temp_user_notes) T

http://sqlfiddle.com/#!9/8aac6a/19

Answer 2

试试这个，似乎有点复杂，但希望对你有帮助;）

Mysql Schema：

CREATE TABLE table1
    (`user_id` int, `category` varchar(10), `suburb` varchar(3), `dated` datetime, `walk_time` int)
;

INSERT INTO table1
    (`user_id`, `category`, `suburb`, `dated`, `walk_time`)
VALUES
    (1, 'experience', 'US', '2016-04-09 00:00:00', 5),
    (1, 'discovery', 'US', '2016-04-09 00:00:00', 5),
    (1, 'experience', 'UK', '2016-04-09 00:00:00', 5),
    (1, 'experience', 'AUS', '2016-04-23 00:00:00', 10),
    (2, 'actions', 'IND', '2016-04-15 00:00:00', 2),
    (2, 'actions', 'IND', '2016-04-15 00:00:00', 1),
    (2, 'discovery', 'US', '2016-04-21 00:00:00', 2),
    (3, 'discovery', 'FR', '2016-04-12 00:00:00', 3),
    (3, 'Emotions', 'IND', '2016-04-23 00:00:00', 3),
    (3, 'discovery', 'UK', '2016-04-12 00:00:00', 4),
    (3, 'experience', 'IND', '2016-04-12 00:00:00', 3)
;

查询SQL：

select c.user_id, c.category, s.suburb, d.dated, w.walk_time
from (
    select user_id, left(group_concat(category order by cnt desc), locate(',', group_concat(category order by cnt desc)) - 1) as category
    from (
        select
            user_id, category, count(1) as cnt
        from table1
        group by user_id, category
    ) t
    group by user_id
) c
inner join (
    select user_id, left(group_concat(suburb order by cnt desc), locate(',', group_concat(suburb order by cnt desc)) - 1) as suburb
    from (
        select
            user_id, suburb, count(1) as cnt
        from table1
        group by user_id, suburb
    ) t
    group by user_id
) s on c.user_id = s.user_id
inner join (
    select user_id, left(group_concat(dated order by cnt desc), locate(',', group_concat(dated order by cnt desc)) - 1) as dated
    from (
        select
            user_id, dated, count(1) as cnt
        from table1
        group by user_id, dated
    ) t
    group by user_id
) d on c.user_id = d.user_id
inner join (
    select user_id, left(group_concat(walk_time order by cnt desc), locate(',', group_concat(walk_time order by cnt desc)) - 1) as walk_time
    from (
        select
            user_id, walk_time, count(1) as cnt
        from table1
        group by user_id, walk_time
    ) t
    group by user_id
) w on c.user_id = w.user_id

<强>结果：

| user_id |  category  | suburb |        dated        | walk_time | 
+---------+------------+--------+---------------------+-----------+
| 1       | experience | US     | 2016-04-09 00:00:00 | 5         | 
| 2       | actions    | IND    | 2016-04-15 00:00:00 | 2         | 
| 3       | discovery  | IND    | 2016-04-12 00:00:00 | 3         |

如何在mysql中为每个用户获取最多的行

2 个答案: